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Jordan Canonical Form

RAJEEV SINGH

December 4th, 2006

In this article I shall tell the algorithm to find the Jordan Canonical Form of a given square matrix. ThenI shall give an example to illustrate the method. I shall not give any formal mathematical proof for any ofthe claims which I make but the proofs should be looked up from any textbook on matrix analysis.

Algorithm

Find all the eigenvalues of the given n n square matrix A. Let the eigenvalues be denoted by (m)i

, wherem denotes the degeneracy of the ith eigenvalue. Find all the independent vectors satisfying the equation

A

(m)i Ix= 0

where I is n n identity matrix and xis n-dimensional column vector. All the vectors satisfying the aboveequations are called rank 1eigenvectors ofA. If we are able to find n independent rank-1 eigenvectors ofAthen we would be able to diagonalize it. We form a nnmatrix (sayR) with the independent eigenvectorsby placing them as columns. Then we find R1. The matrix

RAR1

gives the diagonal form ofA.

If we get lesser number of independent vectors, then we need to find the independent vectors satisfying theequation

A (m)i I2

x= 0

The vectors satisfying this equation and not satisfying the previous equation are called rank 2eigenvectorsofA. We then form chains from rank-2 eigenvetors by operating them by

A

(m)i I

. Then we see if we

have foundn independent vectors including all the rank-1 and rank-2 chain of eigenvectors. Again if we havefound n independent vectors, we form the matrix R and find the product

RAR1

, which gives the Jordan

Canonical Form ofA.Again if we get lesser number of independent vectors, we find the independent vectors satisfying the equation

A

(m)i I3

x= 0

The vectors satisfying this equation and not satisfying the earlier two equations are called rank 3 eigen-

vectors ofA. Again we find chains by successively operating byA

(m)i I

, and if we succeed in finding

n independent vectors we form the matrix R and find the Jordan Canonical Form using it.We repeat the above steps to find rank-4, rank-5, etc. eigenvectors ofA till we find n independent of them.Then we form the matrix R and use it to find the Jordan Canonical Form ofA.

Next we give an illustrative example giving each step of calculation which would explain the method justoutlined.

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Example

Let the given square matrix be

A=

1

2

10 24 18 54 121 6 3 6 1

1 8 0 16 41 0 2 2 01 4 0 6 2

All the eigenvalues of this matrix are 2, i.e. it has a five-fold degeneracy in its spectrum.

A 2I= 1

2

6 24 18 54 121 2 3 6 1

1 8 4 16 41 0 2 2 01 4 0 6 2

Rank-1 eigenvectors are given by

(A 2I)

x1x2x3x4x5

= 0

1

2

6 24 18 54 121 2 3 6 1

1 8 4 16 41 0 2 2 01 4 0 6 2

x1x2x3x4x5

= 0

Using Gauss elimination method to solve these equations,

6 24 18 54 121 2 3 6 1

1 8 4 16 41 0 2 2 01 4 0 6 2

1 4 3 9 21 2 3 6 1

1 8 4 16 41 0 2 2 01 4 0 6 2

1 4 3 9 20 2 0 3 10 4 1 7 20 4 1 7 20 8 3 15 4

1 4 3 9 20 2 0 3 10 0 1 1 00 0 1 1 00 0 3 3 0

1 4 3 9 20 2 0 3 10 0 1 1 00 0 0 0 00 0 0 0 0

Thus we get three equations,

x3+ x4= 0 x3= x4

2x2+ 3x4 x5= 0 x5= 2x2+ 3x4

x1 4x2 3x3+ 9x4 2x5= 0

x1 4x2 3x4+ 9x4 2 (2x2+ 3x4) = 0 x1= 0

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Thus we have two independent rank-1 eigenvectors which are

0100

2

and

00113

.

Since we have just two independent eigenvectors of rank-1, we now need to look for rank-2 eigenvectors,

which are given by

(A 2I)2

x1x2x3x4x5

= 0

Now,

(A 2I)2 =

1

4

6 24 18 54 121 2 3 6 1

1 8 4 16 41 0 2 2 0

1

4 0 6

2

6 24 18 54 121 2 3 6 1

1 8 4 16 41 0 2 2 0

1

4 0 6

2

= 1

4

0 0 0 0 00 0 0 0 0

2 8 6 18 42 8 6 18 46 24 18 54 12

It is easy to see that here we have just one independent equation, which is

x1+ 4x2+ 3x3 9x4+ 2x5= 0

x1= 4x2+ 3x3 9x4+ 2x5

Thus four independent rank-2 eigenvectors are

4

1000

,

3

0100

,

9

0010

and

2

0001

.

The corresponding chains are given by successively operating by (A 2I), and we get

41000

01224

0

30100

1

2

00

113

0

90010

1

2

0377

15

0

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20001

1

2

01224

0

One can check that out of all the generalized eigenvectors we have found so far only four are independent,so we need to find rank-3 eigenvectors, which are given by

(A 2I)3

x1x2x3x4x5

= 0

Now,

(A 2I)3

=

1

8

0 0 0 0 00 0 0 0 0

2 8 6 18 42 8 6 18 46 24 18 54 12

6 24 18 54 121 2 3 6 1

1 8 4 16 41 0 2 2 01 4 0 6 2

= 0

Hence the rank-3 eigenvectors can be taken as

10000

,

01000

,

00100

,

00010

and

00001

.

The corresponding chains are given by successively operating by (A 2I), and we get

1

0000

1

2

6

11

11

1

2

0

0113

0

01000

121

40

2

00226

0

001

00

1

2

183

4

20

1

2

003

39

0

00010

273

81

3

1

2

00

9927

0

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00001

1

2

121

40

2

00113

0

Again one can check that it is not possible to find five independent eigenvectors with a rank-3 chain and anytwo rank-1 eigenvectors. Hence one has to look for a rank-3 and a rank-2 chain. There are many possiblecombinations of independent eigenvectors, out of which we choose the first chain of rank-3 and the first one of

rank-2, and the five independent eigenvectors are

10000

, 1

2

61

111

, 1

2

00

113

,

41000

and

01224

.

We form the following matrix using these

R=1

2

2 6 0 8 00 1 0 2 20 1 1 0 4

0 1 1 0 40 1 3 0 8

Its inverse is

R1 = 1

2

2 6 0 8 00 1 0 2 20 1 1 0 40 1 1 0 40 1 3 0 8

The Jordan Canonical Form is given by

RAR1 =

2 0 0 0 01 2 0 0 00 1 2 0 00 0 0 2 00 0 0 1 2

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